4-fluoro-2-methyl-1H-indol-5-ol → HCID 10162785, compound

HRID 454223Notes

Reaction equation

HRID 454223 reaction equation
Cl[C:2]1[C:11]2[C:6](=[CH:7][C:8]([O:14][CH2:15][CH2:16][CH2:17][N:18]3[CH2:23][CH2:22][N:21]([CH3:24])[CH2:20][CH2:19]3)=[C:9]([O:12][CH3:13])[CH:10]=2)[N:5]=[CH:4][N:3]=1.[F:25][C:26]1[C:34]([OH:35])=[CH:33][CH:32]=[C:31]2[C:27]=1[CH:28]=[C:29]([CH3:36])[NH:30]2>>[F:25][C:26]1[C:34]([O:35][C:2]2[C:11]3[C:6](=[CH:7][C:8]([O:14][CH2:15][CH2:16][CH2:17][N:18]4[CH2:23][CH2:22][N:21]([CH3:24])[CH2:20][CH2:19]4)=[C:9]([O:12][CH3:13])[CH:10]=3)[N:5]=[CH:4][N:3]=2)=[CH:33][CH:32]=[C:31]2[C:27]=1[CH:28]=[C:29]([CH3:36])[NH:30]2

Participants

3

Reactant 2

HCID 103520364-fluoro-2-methyl-1H-indol-5-olHCID 112450614-Chloro-6-methoxy-7-(3-(4-methylpiperazin-1-yl)propoxy)quinazoline

Product 1

HCID 10162785HCID 10162785, compoundYield 70%

Procedure

Using an analogous procedure to that described in Example 237, 4-chloro-6-methoxy-7-(3-(4-methylpiperazin-1-yl)propoxy)quinazoline (106 mg, 0.30 mmol), (prepared as described for the starting material in Example 176), was reacted with 4-fluoro-5-hydroxy-2-methylindole (60 mg, 0.36 mmol), (prepared as described for the starting material in Example 237), to give 4-(4-fluoro-2-methylindol-5-yloxy)-6-methoxy-7-(3-(4-methylpiperazin-1-yl)propoxy)quinazoline (100 mg, 70%).

Workup

  1. custom(prepared
  2. custom(prepared

Sources & evidence

ORD record
ord-8682e682fdb645eabe5224a1733d9ffc
Source dataset
uspto-grants-2013_07
Patent number
US08492560B2

Reaction notes

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